$f(n)=-48\cdot\left(-\dfrac{1}{4}\right)^{{\,n}}$ Complete the recursive formula of $f(n)$. $f(1)=$
Answer: $f( 1)=-48\cdot\left(-\dfrac14\right)^{ 1}={12}$ $f( 2)=-48\cdot\left(-\dfrac14\right)^{ 2}={-3}$ $\dfrac{f( 2)}{f( 1)}=\dfrac{{-3}}{{12}}={-\dfrac14}$ So the first term of the sequence is ${12}$ and the common difference is ${-\dfrac14}$. This is the recursive formula of the sequence: $\begin{cases} f(1)={12} \\\\ f(n)=f(n-1)\cdot\left({-\dfrac14}\right) \end{cases}$